Stoichiometry example problem 1

We know that solid phosphorus will react with chlorine gas to spontaneously produce phosphorus trichloride, liquid phosphorus trichloride. And we're told that we have 1.45 grams of solid molecular phosphorus. And we're asked how many grams of chlorine is required to essentially use up all of the phosphorus that we have? And how many grams of phosphorus trichloride is going to be produced? Now before you do any of these stoichiometry problems. And that's just a fancy word for problems where you need to figure out how much of a certain reactant is required. Or how much of a product is going to be produced. Before you do any of these problems you have to make sure that your reaction, or that your equation, is balanced. So let's make sure. So on the left-hand side, here, this molecule of phosphorus has four phosphorus atoms. So on the whole left-hand side, all of our reactants combined have 4 phosphorus atoms. So our products also need to have 4 phosphorus atoms. But the way it's written right now, I only have 1. So let me just multiply this guy by 4. Now I have 4 phosphorus atoms on both sides of the equation. Let's balance the chlorine now. On the left-hand side I only have 2 chlorine atoms. This 1 molecule of chlorine has 2 atoms in it. Here I have each molecule of phosphorus trichloride has 3 chlorines. And I have 4 molecules of it. So 4 times 3, I have 12 chlorine's on the right-hand side. So 12 on the right-hand side. I need to have 12 on the left. I only have 2 here. Let me multiply that by 6. 6 times 2 is 12. 4 times 3 is 12. Now our equation is all balanced. 4 phosphoruses each side, and twelve chlorines. Now the next thing we have to do, now that we know that we have a balanced equation and we can get into the meat of the problem, is figure out how many moles of phosphorus we're dealing with. Because once we know the moles, we can use stoichiometric ratios. Which is just essentially saying, look, for every mole of that, I need 6 moles of that. And for every mole of that, I'm going to produce 4 moles of that. So you want to get it all in terms of moles. So let's figure out how many moles of phosphorus we have on our hands. Let's look at our periodic table. This periodic table, we have to give proper attribution to the maker. It's made by Le Van Han Cedric. I got this off of Wikimedia Creative Commons. It had an attribution license. Other than that, we're free to use it. And let's go to phosphorus. Phosphorous right here has its atomic weight of 30.974. Let's just round that up to 31. Let me write it down right here. So phosphorus has atomic weight of 31. Which tells us that a mole of phosphorus will weigh 31 grams. Remember a mole is this, you know, 6.02 times 10 to the 20th. It's this huge number of atoms. If you have that many number of atoms of atomic phosphorous. It's going to weight at 31 grams. Now if you look at the atomic weight of P4, or a molecule that has 4 phosphorus atoms in it, it's going to be 4 times this. So it's going to have an atomic weight of-- well, what's 4 times 31? it's 124-- of 124. Which means that 1 mole of-- let me write it here-- this tells us, right there, that 1 mole of solid molecular phosphorus is going to have a-- well, since we're doing it in grams, I should say, will have a mass of 124 grams. Now, given that, we can use that with this information to figure out how many moles of molecular phosphorus we have. Solid molecular phosphorus. So let me start over here. So we have 1.45 grams-- let me write it this way-- of molecular phosphorus, each molecule containing 4 atoms. And we could just do a little dimensional analysis, and make sure everything cancels out. Times this information right here. We have 1 mole of molecular phosphorus for every 124 grams of molecular phosphorus. I should write this here, just so we remember what we're talking about. 1 mole of molecular phosphorus for every 124 grams of molecular phosphorus. And then this cancels out. So the units at least cancel out. The grams of phosphorus. And then we get 1.45 times 1/124 moles of molecular phosphorus. And we can figure that out. Let's get our calculator out. It's 1.45-- we could just say divided by 124, right? That's the same thing as multiplying by 1/124. So divided by 124 is equal to 0.001-- I'll just round it, 7. So this is equal to 0.0117 moles of phosphorus. That's what we're starting off with. That's what this 1.45 grams are. And that makes sense. If we had an entire mole of molecular phosphorus, it would weigh a 124 grams. We only have 1.45 of that. So it's a little bit more than 1/100, which makes sense. This number right here is a little more than 1/100. Now, we need to think about, for every mole, let's scroll down here. Although I don't want to lose my equation. We need to think about for every mole-- so let me write down, so we have 0.0117 moles of molecular phosphorus. And for every mole of phosphorus, how many moles of chlorine molecules do we need? So let me write this down. So for every 6 moles of chlorine gas-- I'll do it in blue-- for every 6 moles of chlorine gas, we need 1 mole of molecular phosphorus. And the reason I wrote it this way, instead of writing 1 mole of molecular phosphorus for every 6 moles of chlorine gas, is because I want to make sure the units cancel out. And it also should make sense for you intuitively. If I have 1 of this, I'm going to need 6 of this. If I have 0.0117 of this, I'm going to need 6 times of the chlorine gas. And the units work out. This cancels out with that. And I'm just essentially multiplying by 6. So we're going to have 6 times 0.0117. I think I still have that on my calculator. Yes, it's right there. So times 6, is equal to 0.07. I'll just keep it there because then it is 0 again. So this is equal to 0.07 moles of chlorine gas required. And I could write it right here. Let me write, required. Required in the numerator. Required in the denomenator. That makes a little bit more sense or it clarifies things. For every 6 moles of chlorine gas that are required, 1 mole of phosphorus is required, of solid phosphorus, is required. Or we could say for every mole of phosphorus is required, you need 6 moles of chlorine gas. We got that just from the original equation. So I'll write this, required, down right there. So we're almost there. We figured out that 0.07 moles of chlorine gas are required. But we want to find out how many grams of chlorine gas are required. So we just have to figure out how many grams there are per mole? So let's figure that out. Let's look at the chlorine. Chlorine is off here to the right. This is a super huge periodic table. Chlorine, right here, has an atomic weight of 35.453. So chlorine, 35.453 atomic weight. The weighted average of all of the isotopes of chlorine on earth. So molecular chlorine gas, which has 2 atoms, is going to have twice that. So I could do it in my head. It is going to be, right, about what? 70.9? I'll just use a calculator. I don't want to make a silly math mistake. 35.453 times 2 is equal to 70.906. I didn't even need the calculator. So it's equal to 70.906. My memory is bad. 70.906 atomic weight. Which tells us that-- well, let me write this information that we had down before. We were required to have 0.07 moles of chlorine gas. That is what is required. And let's multiply it. We want moles in the denominator. So we want moles in the denominator here. So 1 mole of chlorine gas. For every mole, what's the mass going to be? Well if the atomic weight of chlorine gas is 70.906, that means one mole of it is going to have a mass of 70.906 grams. So for every mole, we have 70.90 grams. We have 0.07 moles. So we will multiply 0.07 times 70 to figure out how many grams we have. And the units cancel out. We have moles of chlorine gas and we're just going to have grams required. So this is going to be equal to 70 times 0.07 grams required. So let's get that over here. So we already have the 70.906 times 0.07 is equal to 4.96 grams. So 4.96 grams of-- and actually this is grams of chlorine gas. I should write that there. So grams of chlorine gas are required. Even the required worked out with the dimensional analysis. And we've answered the first part of the problem. If we have 1.45 grams of phosphorus as one of the reactants, then we're going to need 4.96. 4.96 grams of chlorine are required. Now, the second question is how much phosphorus trichloride is produced? Now there's two ways to do it. An easy way and a hard way. The easy way says, well, mass is conserved. If you have grams on the side, the total mass on this side has to be equal to the total mass on that side. That's the easy way. The slightly harder way is you could have done the exact same thing we did with the chlorine gas, you could do it with the phosphorus trichloride. You could say, hey, for every mole of this, I need 4 moles of that. And then figure out the mass of each mole and multiply and do all of that. But let's do it the easy way. We know that we have 1.45 grams of this reactant, and we have 4.96 grams of this reactant. So this, which is the only product right here, this product right here must have a mass of our combined reactants, because it's going to react fully. So let's figure that out. We have 4.96. I'll just round 4.96 plus 1.45 is equal to 6.41 grams of phosphorus trichloride. 6.41 grams. And we're done.